What will come in place of question mark (?) in the following questions
`4 (4)/(5) + 2 (1)/(15) - (27)/(5) = 2 (1)/(5) div 3 xx ?`

Options

A. `(2)/(9)`
B. 1
C. 2
D. 3

Answer

To solve the equation \( 4 \frac{4}{5} + 2 \frac{1}{15} - \frac{27}{5} = 2 \frac{1}{5} \div 3 \times ? \), we will follow these steps:

### Step 1: Convert Mixed Numbers to Improper Fractions
Convert \( 4 \frac{4}{5} \) and \( 2 \frac{1}{15} \) into improper fractions.

- For \( 4 \frac{4}{5} \):
\[ 4 \frac{4}{5} = \frac{4 \times 5 + 4}{5} = \frac{20 + 4}{5} = \frac{24}{5} \]

- For \( 2 \frac{1}{15} \):
\[ 2 \frac{1}{15} = \frac{2 \times 15 + 1}{15} = \frac{30 + 1}{15} = \frac{31}{15} \]

- For \( 2 \frac{1}{5} \):
\[ 2 \frac{1}{5} = \frac{2 \times 5 + 1}{5} = \frac{10 + 1}{5} = \frac{11}{5} \]

### Step 2: Rewrite the Equation
Now, rewrite the equation using the improper fractions:
\[ \frac{24}{5} + \frac{31}{15} - \frac{27}{5} = \frac{11}{5} \div 3 \times ? \]

### Step 3: Find a Common Denominator
To add and subtract the fractions on the left side, we need a common denominator. The least common multiple (LCM) of 5 and 15 is 15.

- Convert \( \frac{24}{5} \) to have a denominator of 15:
\[ \frac{24}{5} = \frac{24 \times 3}{5 \times 3} = \frac{72}{15} \]

- Convert \( \frac{27}{5} \) to have a denominator of 15:
\[ \frac{27}{5} = \frac{27 \times 3}{5 \times 3} = \frac{81}{15} \]

### Step 4: Substitute and Simplify
Now substitute back into the equation:
\[ \frac{72}{15} + \frac{31}{15} - \frac{81}{15} = \frac{11}{5} \div 3 \times ? \]

Combine the fractions on the left side:
\[ \frac{72 + 31 - 81}{15} = \frac{22}{15} \]

### Step 5: Simplify the Right Side
Now simplify the right side:
\[ \frac{11}{5} \div 3 = \frac{11}{5} \times \frac{1}{3} = \frac{11}{15} \]
Thus, the equation becomes:
\[ \frac{22}{15} = \frac{11}{15} \times ? \]

### Step 6: Solve for ?
To find \( ? \), we can multiply both sides by \( \frac{15}{11} \):
\[ ? = \frac{22}{15} \times \frac{15}{11} = \frac{22}{11} = 2 \]

### Final Answer
Therefore, the value that comes in place of the question mark (?) is:
\[ \boxed{2} \]